đề bài sai thánh giải đc à

a) Q=13-(x^2+4x+4)=13-(x+2)^2<=13 Q_max=13 khi x=-2

b) M=\(6x-x^2+74+x=74-\left(x^2+7x\right)=74-\left(x^2-2.\frac{7}{2}x+\left(\frac{7}{2}\right)^2\right)^{^2}-\left(\frac{7}{2}\right)^2\\ \)

\(\frac{74\cdot4-49}{4}-\left(x-\frac{7}{2}\right)^2\le\frac{74\cdot4-49}{4}=M_{max}\)đảng thức khi x=7/2

C) \(P=\frac{25}{4}-\left(x^2-2.\frac{5}{2}+\left(\frac{5}{2}\right)^2\right)=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}=P_{max}\) khi x=5/2

( a+ b+c) ^2 >= 3(ab+ bc+ ca)

=> a^2 + b^2 + c^2 + 2ab+ 2bc + 2ac >= 3ab + 3bc + 3ac

=> a^2 + b^2 + c^2 + 2ab + 2bc+ 2ac - 3ab - 3bc - 3ac >=0

=> a^2 + b^2 + c^2 - ab- bc - ac >=0

=> 2( a^2 + b^2 + c^2 - ab-bc-ac) >=0

=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac >=0

=> a^2 - 2ab + b^2 + b^2 -2bc + c^2 + c^2 - 2ac + a^2 >=0

=> (a-b)^2 +(b-c)^2 +(c-a)^2 >=0

vì (a-b)^2 >=0

(b-c)^2 >=0

(c-a)^2 >=0

nên (a-b)^2 +(b-c)^2 +(c-a)^2 >=0

hay (a+b+c)^2 >= 3(ab+bc+ca)

= 3 nha

1/a + 1/b + 1/c = 1/a+b+c

=> (  ab + bc + ca ) x ( a + b +c ) = abc 

=> ( ab + bc + ca ) x ( a + b ) + ( abc + bcc + cca - abc ) = 0 

=> ( ab + bc + ca ) x ( a + b ) + c²  x ( a + b ) = 0

=> ( a + b ) x ( a + c ) x ( b + c ) = 0

=> trong đó a , b đối nhau khi đó vì n lẻ nên

1/a²⁰¹³ + 1/b²⁰¹³ + 1/c²⁰¹³ = 1/c²⁰¹³ = 1/c²⁰¹³ + b ²⁰¹³ + c²⁰¹³

cau xem fairy tail a

Lời giải :

Đặt \(\frac{a}{b}=t\Leftrightarrow\frac{b}{a}=\frac{1}{t}\)

BĐT \(\Leftrightarrow t^2+\frac{1}{t^2}+4\ge3\left(t+\frac{1}{t}\right)\)

\(\Leftrightarrow\left(t+\frac{1}{t}\right)^2-3\left(t+\frac{1}{t}\right)+2\ge0\)

\(\Leftrightarrow\left(t+\frac{1}{t}-1\right)\left(t+\frac{1}{t}-2\right)\ge0\)

\(\Leftrightarrow\frac{t^2-t+1}{t}\cdot\frac{t^2-2t+1}{t}\ge0\)

\(\Leftrightarrow\frac{\left(t^2-t+1\right)\left(t-1\right)^2}{t^2}\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow t=1\Leftrightarrow\frac{a}{b}=1\Leftrightarrow a=b\)

\(\frac{x^2-x+1}{x}=5\)

\(\Leftrightarrow x-1+\frac{1}{x}=5\)

\(\Leftrightarrow x+\frac{1}{x}=6\)

\(\Leftrightarrow x^2+\frac{1}{x^2}=34\)

Mà ta lại có

\(M=\frac{x^4+x^2+1}{x^2}=x^2+1+\frac{1}{x^2}\)

\(=34+1=35\)